Dustinsfl said:So for a force to be conservative it can only depend on position and the work as to be the same for all paths.
The force mass time gravity is conservative but how do I show the all paths are the same?
Dustinsfl said:Since gravity is acting along the radius to the center of earth, then would ##\mathbf{F}(\mathbf{r}) = -m\mathbf{r}##?
Dick said:That is a conservative force, but I don't think Newton would agree it has much to do with gravity. What are you thinking?
Dustinsfl said:I wrote why I put that.
I said since gravity is acting along the raidus vector to the center of Earth then could I wrtie F in such a manner.
A conservative force is a type of force that does work on an object that is independent of the path taken by the object. This means that the work done by the force is the same regardless of the path taken, as long as the starting and ending points are the same.
It is important to prove that work is the same for all paths of a conservative force because it helps us understand and predict the behavior of objects under the influence of these forces. It also allows us to simplify calculations and make more accurate predictions in various fields of science and engineering.
We can use the fundamental theorem of calculus, which states that the integral of a function between two points is equal to the difference of the function evaluated at those points. In this case, we can use this theorem to show that the work done by a conservative force is equal to the difference in potential energy between the starting and ending points, which is independent of the path taken.
Some common examples of conservative forces include gravity, electric forces, and magnetic forces. These forces all follow the principle of work being independent of the path taken and have corresponding potential energy functions.
No, a non-conservative force does not have the same work for all paths. This is because non-conservative forces, such as friction or air resistance, depend on the path taken by the object and dissipate energy as heat. Therefore, the work done by these forces is not independent of the path taken.